Using vectors to find distances

7.8 USING VECTORS TO FIND DISTANCES
of the sphere. This is easily expressed in vector notation as
|r − c|2 = (r − c) · (r − c) = a2 ,
(7.43)
where c is the position vector of the centre of the sphere and a is its radius.
Find the radius ρ of the circle that is the intersection of the plane n̂ · r = p and the sphere
of radius a centred on the point with position vector c.
The equation of the sphere is
|r − c|2 = a2 ,
(7.44)
|r − b|2 = ρ2 ,
(7.45)
and that of the circle of intersection is
where r is restricted to lie in the plane and b is the position of the circle’s centre.
As b lies on the plane whose normal is n̂, the vector b − c must be parallel to n̂, i.e.
b − c = λn̂ for some λ. Further, by Pythagoras, we must have ρ2 + |b − c|2 = a2 . Thus
λ2 = a2 − ρ 2 .
Writing b = c + a2 − ρ2 n̂ and substituting in (7.45) gives
r 2 − 2r · c + a2 − ρ2 n̂ + c2 + 2(c · n̂) a2 − ρ2 + a2 − ρ2 = ρ2 ,
whilst, on expansion, (7.44) becomes
r2 − 2r · c + c2 = a2 .
Subtracting these last two equations, using n̂ · r = p and simplifying yields
p − c · n̂ = a2 − ρ2 .
On rearrangement, this gives ρ as a2 − (p − c · n̂)2 , which places obvious geometrical
constraints on the values a, c, n̂ and p can take if a real intersection between the sphere
and the plane is to occur. 7.8 Using vectors to find distances
This section deals with the practical application of vectors to finding distances.
Some of these problems are extremely cumbersome in component form, but they
all reduce to neat solutions when general vectors, with no explicit basis set,
are used. These examples show the power of vectors in simplifying geometrical
problems.
7.8.1 Distance from a point to a line
Figure 7.14 shows a line having direction b that passes through a point A whose
position vector is a. To find the minimum distance d of the line from a point P
whose position vector is p, we must solve the right-angled triangle shown. We see
that d = |p − a| sin θ; so, from the definition of the vector product, it follows that
d = |(p − a) × b̂|.
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VECTOR ALGEBRA
P
p−a
d
p
θ
A
b
a
O
Figure 7.14 The minimum distance from a point to a line.
Find the minimum distance from the point P with coordinates (1, 2, 1) to the line r = a+λb,
where a = i + j + k and b = 2i − j + 3k.
Comparison with (7.39) shows that the line passes through the point (1, 1, 1) and has
direction 2i − j + 3k. The unit vector in this direction is
1
b̂ = √ (2i − j + 3k).
14
The position vector of P is p = i + 2j + k and we find
1
(p − a) × b̂ = √ [ j × (2i − 3j + 3k)]
14
1
= √ (3i − 2k).
14
Thus the minimum distance from the line to the point P is d =
13/14. 7.8.2 Distance from a point to a plane
The minimum distance d from a point P whose position vector is p to the plane
defined by (r − a) · n̂ = 0 may be deduced by finding any vector from P to the
plane and then determining its component in the normal direction. This is shown
in figure 7.15. Consider the vector a − p, which is a particular vector from P to
the plane. Its component normal to the plane, and hence its distance from the
plane, is given by
d = (a − p) · n̂,
where the sign of d depends on which side of the plane P is situated.
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(7.46)
7.8 USING VECTORS TO FIND DISTANCES
P
n̂
d
p
a
O
Figure 7.15 The minimum distance d from a point to a plane.
Find the distance from the point P with coordinates (1, 2, 3) to the plane that contains the
points A, B and C having coordinates (0, 1, 0), (2, 3, 1) and (5, 7, 2).
Let us denote the position vectors of the points A, B, C by a, b, c. Two vectors in the
plane are
b − a = 2i + 2j + k
and
c − a = 5i + 6j + 2k,
and hence a vector normal to the plane is
n = (2i + 2j + k) × (5i + 6j + 2k) = −2i + j + 2k,
and its unit normal is
n̂ =
n
= 13 (−2i + j + 2k).
|n|
Denoting the position vector of P by p, the minimum distance from the plane to P is
given by
d = (a − p) · n̂
= (−i − j − 3k) · 13 (−2i + j + 2k)
=
2
3
−
1
3
− 2 = − 53 .
If we take P to be the origin O, then we find d = 13 , i.e. a positive quantity. It follows from
this that the original point P with coordinates (1, 2, 3), for which d was negative, is on the
opposite side of the plane from the origin. 7.8.3 Distance from a line to a line
Consider two lines in the directions a and b, as shown in figure 7.16. Since a × b
is by definition perpendicular to both a and b, the unit vector normal to both
these lines is
a×b
.
n̂ =
|a × b|
231
VECTOR ALGEBRA
b
Q
q
n̂
P
p
a
O
Figure 7.16 The minimum distance from one line to another.
If p and q are the position vectors of any two points P and Q on different lines
then the vector connecting them is p − q. Thus, the minimum distance d between
the lines is this vector’s component along the unit normal, i.e.
d = |(p − q) · n̂|.
A line is inclined at equal angles to the x-, y- and z-axes and passes through the origin.
Another line passes through the points (1, 2, 4) and (0, 0, 1). Find the minimum distance
between the two lines.
The first line is given by
r1 = λ(i + j + k),
and the second by
r2 = k + µ(i + 2j + 3k).
Hence a vector normal to both lines is
n = (i + j + k) × (i + 2j + 3k) = i − 2j + k,
and the unit normal is
1
n̂ = √ (i − 2j + k).
6
A vector between the two lines is, for example, the one connecting the points (0, 0, 0)
and (0, 0, 1), which is simply k. Thus it follows that the minimum distance between the
two lines is
1
1
d = √ |k · (i − 2j + k)| = √ . 6
6
7.8.4 Distance from a line to a plane
Let us consider the line r = a + λb. This line will intersect any plane to which it
is not parallel. Thus, if a plane has a normal n̂ then the minimum distance from
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